Any literal suffix appearing twice in a term is a dummy suffix, which may be changed freely to any other letter not already appropriated in that term. Two or more dummy suffixes can be interchanged\footnotemark.\footnotetext {At first we shall call attention to such changes when we employ them; but the reader will be expected gradually to become familiar with the device as a common process of manipulation.} For example \index{Dummy suffixes}% \[ g_{\alpha\beta}\, \frac{\dd^{2} x_{\alpha}}{\dd x_{\mu}'\, \dd x_{\nu}'}\, \frac{\dd x_{\beta}}{\dd x_{\lambda}'} = g_{\alpha\beta}\, \frac{\dd^{2} x_{\beta}}{\dd x_{\mu}'\, \dd x_{\nu}'}\, \frac{\dd x_{\alpha}}{\dd x_{\lambda}'} \Tag{(22.2)} \] {\Loosen by interchanging the dummy suffixes $\alpha$ and~$\beta$, remembering that $g_{\beta\alpha} = g_{\alpha\beta}$.} For a further illustration we shall prove that \[ \left. \begin{aligned} \frac{\dd x_{\mu}}{\dd x_{\alpha}'}\, \frac{\dd x_{\alpha}'}{\dd x_{\nu}} = \frac{\dd x_{\mu}}{\dd x_{\nu}} &= 0,\quad \text{if $\mu \neq \nu$} \\ &= 1,\quad \text{if $\mu = \nu$} \end{aligned} \right\}. \Tag{(22.3)} \] The left-hand side written in full is \[ \frac{\dd x_{\mu}}{\dd x_{1}'}\, \frac{\dd x_{1}'}{\dd x_{\nu}} + \frac{\dd x_{\mu}}{\dd x_{2}'}\, \frac{\dd x_{2}'}{\dd x_{\nu}} + \frac{\dd x_{\mu}}{\dd x_{3}'}\, \frac{\dd x_{3}'}{\dd x_{\nu}} + \frac{\dd x_{\mu}}{\dd x_{4}'}\, \frac{\dd x_{4}'}{\dd x_{\nu}}, \] which by the usual theory gives the change~$dx_{\mu}$ consequent on a change~$dx_{\nu}$. But $x_{\mu}$~and $x_{\nu}$ are coordinates of the same system, so that their variations are independent; hence $dx_{\mu}$~is zero unless $x_{\mu}$~and $x_{\nu}$ are the same coordinate, in which case, of course, $dx_{\mu} = dx_{\nu}$. Thus the theorem is proved. The multiplier $\dfrac{\dd x_{\mu}}{\dd x_{\alpha}'}\, \dfrac{\dd x_{\alpha}'}{\dd x_{\nu}}$ acts as a \emph{substitution-operator}. That is to say if \index{Substitution-operator}% $A(\mu)$~is any expression involving the suffix~$\mu$ \[ \frac{\dd x_{\mu}}{\dd x_{\alpha}'}\, \frac{\dd x_{\alpha}'}{\dd x_{\nu}}\, A(\mu) = A(\nu). \Tag{(22.4)} \] For on the left the summation with respect to~$\mu$ gives four terms corresponding to the values $1$,~$2$, $3$, $4$ of~$\mu$. One of these values will agree with~$\nu$. Denote the other three values by $\sigma$, $\tau$,~$\rho$. Then by~\Eq{(22.3)} the result is \begin{align*} 1 · &A(\nu) + 0 · A(\sigma) + 0 · A(\tau) + 0 · A(\rho) \\ {}={} & A(\nu). \end{align*} The multiplier accordingly has the effect of substituting~$\nu$ for~$\mu$ in the multiplicand. \Section{23.}{Tensors} \index{Tensor}% The two laws of transformation given in \SecRef{19} are now written--- Contravariant vectors \begin{align*} A'^{\mu} &= \frac{\dd x_{\mu}'}{\dd x_{\alpha}}\, A^{\alpha}. \Tag{(23.11)} \\ \intertext{\hspace*{\parindent}Covariant vectors} A_{\mu}' &= \frac{\dd x_{\alpha}}{\dd x_{\mu}'}\, A_{\alpha}. \Tag{(23.12)} \end{align*}